Hi, I'm trying to understand the following simple flasher circuit:

http://students.washington.edu/ksf/circuit.JPG
I don't really know what I'm doing here, but I think I have a vague

idea of how it works... I assume you have to take the forward

resistance of the LED into account, because otherwise I don't think

the circuit would do anything.

You are right that you can not make the simplifying assumption that

the LED has a constant voltage drop, independent of current. The

changing voltage drop across the LED is the feedback voltage to the

input of the two stage amplifier.

(Let's say a voltage drop of 2V and

resistance 5 ohms.)

Oops. This is exactly what you cannot assume. This circuit varies

the LED current from almost zero (where the voltage drop will be much

less than 2 volts) to severe overload current (limited only by the

current carrying capability of the second transistor), where the wire

bond resistance and the die resistance will come into play. I can

only guess whether 5 ohms is a good approximation of those, unless I

made a measurement of voltage and current, simultaneously during the

peak, or extrapolated from the LED data sheet.

When the circuit is first turned on, Q1 and Q2

turn on, because there isn't enough voltage across the 2M resistor to

keep Q1 off.

A clearer way to think of this state is that the capacitor can be

assumed to start with no voltage drop across it. If we guess that Q2

will be slammed on (we'll check this later), then it is safe to also

assume that with the battery voltage used up across the series

combination of the LED and Q2, then the voltage across the LED must be

quite a bit more than the .6 volts or so it takes to turn on Q1 via

the 1K resistor. The small current through R2 is not of much

consequence in the presence of this >2 volt LED drop feeding the base

of Q1 through 1K. With Q1 driven on by this current, the capacitor

starts to build up a voltage drop that gradually decreases this base

current. But while the base current flows, Q1 is on, and dumps base

current to Q2, so our original guess that Q2 starts by being blasted

on looks good.

C1 begins charging up (the wrong way!) through the 1K

resistor, approaching a steady-state voltage. (Assuming the power

supply is 4.5V, hFE for both transistors is 100, and the above LED

values, I calculate this to be about -1.4V.)

So far, so good.

However, any capacitor

voltage above -1.3V is enough to keep Q1 off once it is turned off,

because that would produce a voltage over 3.8 (4.5-0.7) across the 2M

resistor. Thus, I'm guessing that the capacitor gets charged to near

-1.4V, Q1 turns off, the capacitor discharges to -1.3, and the cycle

repeats. This would account for the "flasher" behavior, since during

the on cycle, C1 is charging through the smaller 1K resistor and

during the off cycle, it's discharging through the 2M resistor. If

this is correct, then I have some questions:

You generally have it right. As soon as the base drive for Q1 gets

low enough to let Q2 come out of saturation, the resistive and diode

drop of the LED suddenly decreases, and the voltage at both ends of

the capacitor head positive (assuming this happens fast enough that

the voltage across the capacitor does not have time to change much).

This produces a positive feedback situation that snaps Q1 completely

off, and continues to a little reverse bias.

Why is the capacitor oriented the wrong way in the circuit?

It has so little voltage across it that it will probably work in

either orientation. Try it the other way and see if the timing cycle

changes. If it gets faster, it means that leakage current is

interfering less with the slow charge time. I agree that it is

backwards.

When the capacitor is charged between -1.3 and -1.4V, it seems

entirely arbitrary whether the transistors are off or on and the

capacitor is charging or discharging, so what decides when the circuit

changes states?

The capacitor does its fast charge through Q1's base. It does its

slow discharge through R2. Q1's base has to be reverse biased during

this process to isolate the transistor. When the base of Q1 starts to

have enough forward bias to start turning on Q2, the current from R2

has to be gig enough to get the 'switch on' positive feedback going

before all its current has been detoured, or the transistors will just

stall at some slight conduction and the capacitor will discharge no

further (because all of R2's current is going to the base, instead of

the capacitor).

I would add an emitter resistor to Q2 to reduce the peak base current

from Q1 and limit the peak LED current. 10 to 47 ohms ought to work.